Question

As a coffee enthusiast you wish to keep your coffee as hot as possible for as long as possible. You wish to find the rate of heat loss of a typical coffee mug. Assume the mug to be a cylinder which has a lid made out of the same material as the rest of the cylinder, and the thickness of the walls is 1.34 cm. If the coffee in the mug is 79.2 °C and the room temperature is 24.0 °C how much more heat does the mug lose through conduction than from radiated energy. See the hint below for important constants.

Assume that the thermal conductivity of all sides of the mug is 0.850 W/m·C, that the mug has an emissivity of 0.760 and the Stefan-Boltzmann Constant is 5.67 × 10-8 W/m2·K4. Remember to properly convert all units and to include all surfaces of the mug.

Answer 1

Given

   coffee mug with thermal conductiviy k = 0.850 W/m.0C

   mug of cylinder of thickness is t = 1.34*10^-2 m

the temperature of the coffee is T2 = 79.2 0C, T1 = 24.0 0C

   emmisivity is e = 0.760
stefan's constant sigma = 5.67*10^-8 W/m^2*k^4

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first the heat lost due to conduction is

   q1 = k*A*dT/t

substituting the values

   q1 = 0.850*A(79.2-24)/(1.34*10^-2) J

   q1 = 3501.4925*A J

now the heat lost by radiation from Stefan's law of radiation
  
   q2 = A*e*sigma*T^4

can be written as
   q2 = A*e*sigma(T2^4 - T1^4)

   q2 = A*0.760*5.67*10^-8(79.2^4 -24.0^4) J

   q2 = 1.6812*A J

now the difference of the heat lost is q1-q2 = 3501.4925*A - 1.6812*A

   q1-q2 = A( 3501.4925 - 1.6812) J
  
   q1-q2 = 3499.8113 J

   so the most of the heat lost is due to the conduction than radiation

ration q1/q2 = (3501.4925*A)/(1.6812*A) = 2082.7
the heat lost by conduction is 2082.7 times of radiation